Question

Найти массу SO3 образовавшейся при окислении 40 грамм оксида S(VI)

Answer 1

2SO2+O2=2SO3 SO2=64×2коэффициент=128 SO3=80×2коэф=160 128------160 40-------х=50 ответ массаSO3=50 гр

Answer 2

Calculating the Mass of SO3 Formed from the Oxidation of 40 Grams of S(VI)

To calculate the mass of SO3 formed from the oxidation of 40 grams of S(VI), we can use the stoichiometry of the reaction. The balanced chemical equation for the oxidation of S(VI) to SO3 is:

2S(VI) + 3O2 → 2SO3

According to the stoichiometry of the reaction, 2 moles of S(VI) react with 3 moles of O2 to produce 2 moles of SO3.

Given that the molar mass of S(VI) is 32 g/mol and the molar mass of SO3 is 80 g/mol, we can calculate the mass of SO3 formed.

Using the given information and the stoichiometry of the reaction, we can calculate the mass of SO3 formed as follows:

Step 1: Calculate the number of moles of S(VI) in 40 grams: - Number of moles of S(VI) = Mass of S(VI) / Molar mass of S(VI) - Number of moles of S(VI) = 40 g / 32 g/mol - Number of moles of S(VI) = 1.25 moles

Step 2: Use the stoichiometry of the reaction to find the number of moles of SO3 formed: - According to the balanced chemical equation, 2 moles of S(VI) produce 2 moles of SO3. - Therefore, the number of moles of SO3 formed = 1.25 moles * (2 moles SO3 / 2 moles S(VI)) - Number of moles of SO3 formed = 1.25 moles

Step 3: Calculate the mass of SO3 formed: - Mass of SO3 = Number of moles of SO3 * Molar mass of SO3 - Mass of SO3 = 1.25 moles * 80 g/mol - Mass of SO3 = 100 grams

Therefore, the mass of SO3 formed from the oxidation of 40 grams of S(VI) is 100 grams.